Basic Idea: dual of being sober 🍻

In a previous post, I have mentioned that the definition I have adopted for the concept of sober is different from the definition gave on the book where I am learning point-free topology:

📚 Jorge Picado & Aleš Pultr Frames and Locales: topology without points. Springer Basel, Series Frontiers in Mathematics, Vol. 28, xx + 400 pp., 2012 (ISBN: 978-3-0348-0153-9).

There, I said I had seen no reason to define sober different from the way I had defined:

When a topological space $(X, \tau)$ is such that every completely prime filter in $\tau$ is of the form $\tau(x)$ for some $x \in X$, we call it sober.

Now… after thinking a bit… 🧘 I see five different definitions for sober. 🤯

The main purpose of this post is to link all those dots. Please, take a look a the post where I talk about filters of closed sets without sets before proceeding.

Just to make things easier, let’s fix a topological space $X$. Denote by $\tau$ the family of open sets, and by $\mathscr{P}(X)$, the family of subsets of $X$. Also, $L$ is a complete lattice.

Filter, to wannabe prime ideal, to wannabe prime filter

In my last post, we have seen that complements take a filter $\mathscr{F}$ to a wannabe prime ideal, $\mathscr{F}^c$. I say “wannabe”, because it might not actually be an ideal… but if it is a proper ideal, it shall be prime. By polishing this a little, we conclude that complements are a bijection between filters and wannabe prime ideals. By restricting it to prime filters, they are in a bijection with prime ideals.

On the other hand, the order dual of a prime ideal $\mathscr{F}^c$ is the prime filter ${\mathscr{F}^c}^*$. And this is also a bijection.

Disclaimer: I have not really checked all those affirmations I made above. 😅😇

Completely prime in five acts 🎭

Original completely prime filter

Our original definition for a completely prime filter $\mathscr{F}$, is that besides being non-empty and proper (not the whole lattice),

This is a condition that the family, $\tau(x)$ of open neighbouhoods of some point must satisfy:

If the union of a family of open sets contains $x$, it is because at least one member of this family already contains $x$.

Complement prime ideal

Notice that, if $\mathscr{F}$ is completely prime, it is, a fortiori, prime. And therefore, $\mathscr{F}^c$ is a prime ideal. And not only that, the contrapositive of equation \eqref{completely_prime_filter} implies that $\mathscr{F}^c$ is closed by arbitrary joins:

But, for a prime ideal, this is equivalent to $\mathscr{F}^c$ having a maximum:

In fact, if $\mathscr{C} \subset \mathscr{F}^c$, since $\mathscr{F}^c$ is an ideal,

With closed elements

If we look at ${\mathscr{F}^c}^*$, instead, we have just the same, but dual:

The filter $\mathscr{F}$ is completely prime if, and only if, ${\mathscr{F}^c}^*$ is a prime filter with minimum element:

$$\begin{equation} {\bigwedge}_* {\mathscr{F}^c}^* \in {\mathscr{F}^c}^*. \end{equation}$$

With prime open elements

An ideal $\mathscr{I}$ that has a maximum element $W$, is just the ideal generated by $W$:

This ideal is prime if, and only if,

In this case, we shall say that the element $W$ is prime.

This way, the $W$ prime elements are in bijection with the completely prime filters, where the bijection and its inverse are given by

This bijection was used in the demonstration of proposition 1.3.1 in our main reference book:

📚 Jorge Picado & Aleš Pultr Frames and Locales: topology without points. Springer Basel, Series Frontiers in Mathematics, Vol. 28, xx + 400 pp., 2012 (ISBN: 978-3-0348-0153-9).

With $*$-prime closed elements

Just the same way, the $*$-prime (in a dual sense) closed elements are in bijection with the completely prime fields. We are not actually saying anything, because the definition for a proper filter $p^*$ being $*$-prime is: whenever $p$ is prime. That is,

The model for this property is the closed sets of the form $\overline{\{x\}}$. If the union of two closed sets contain $\overline{\{x\}}$, it is because at least one of them contain $x$, and therefore, $\overline{\{x\}}$.

That is, not only $\overline{\{x\}}$ is the smallest closed set that contains $x$. It is also true that $\overline{\{x\}}$ cannot be covered by non-trivial closed sets. By non-trivial, I mean to disallow closed sets that already contain the whole set.

The dual of being sober 🍻

All of the above are equivalent ways of saying $\mathscr{F}$ is completely prime. And therefore, they give us equivalent “dual” ways for deciding whether a topological space $X$ is or not sober.

🍩 🍻 🍩

Figure: Two doughnuts trying to be “not sober” using mugs that, in fact, are isomorphic to the doughnuts themselves!

Let me remind you that I have defined sober as:

A topological space (X, $\tau)$ is sober when all completely prime filters are of the form $\tau(x)$ for some $x \in X$.

$$\begin{equation} \tag{🍻} \label{sober:open_filter} \text{$\mathscr{F}$ completely prime } \Leftrightarrow \exists x \in X,\, \mathscr{F} = \tau(x). \end{equation}$$

Do not forget that $\tau(x)$ is always completely prime. So, the $(\Leftarrow)$ always holds.

Of course, we can use any completely prime filter flavour from before to give an equivalent definition for sober. Let’s do that only for sets.

With prime open sets

Notice that

Then, sober $\eqref{sober:open_filter}$ is:

If $W \in \tau$ is a prime open set, that is, for any open sets $A,B \in \tau$,

$$\begin{equation} A \cap B \subset W \Rightarrow A \subset W \text{ or } B \subset W, \end{equation}$$

then, $W = X \setminus \overline{\{x\}}$ for some $x \in X$.

With $*$-prime closed sets

Notice that

where $\mathscr{c}(x)$ is the family of all closed sets that contain $x$. Then, sober $\eqref{sober:open_filter}$ is:

If $F \subset X$ is a $*$-prime closed set, that is, for any closed sets $P$ and $Q$,

$$\begin{equation} F \subset P \cup Q \Rightarrow F \subset P \text{ or } F \subset Q, \end{equation}$$

then, $F = \overline{\{x\}}$ for some $x \in X$.

Better naming convention

It seems that for those open sets above, people do not call them prime. They are called meet-irreducible. And for what I have called $*$-prime, people use join-irreducible.

I really find this “correct” way much better than mine! It does not privilege one over the other. It does not favour open over closed. It is not filter centred, nor ideal centred.

Conclusion

🥳 🍻 😎

Cheers!!!

PS: I do not drink any alcohol, myself.

Change Log

2025-06-15: I have just realized that meet-irreducible and join-irreducible are much better naming.